All local cohomology modules being zero

Question

Let $R$ be a Noetherian ring with unit, $I$ be an ideal of $R$ and let $M$ be a finitely generated $R$-module. Suppose $H_{I}^j(M)=0$ for all $j$, then how can one show that $M=IM$?

The converse of the above is trivial.

Answer 1

This one is easy, too. Assume that $IM\neq M$. Then $\operatorname{grade}(I,M)=n<\infty$. Now prove by induction on $n$ that $\operatorname{grade}(I,M)=\min\{i:H_I^i(M)\neq 0\}$ and get a contradiction.